J-15 Carrier Multirole Fighter thread

Bhurki

Junior Member
Registered Member
Why are you assuming (140 + 47) m and not (68 + 47) m as the launch point distance? What is the end speed based on MTOW of 29 tons?
Thats what they have calculated against, thats the max Mtow any jet is allowed from the aft launch point. Also i think the shorter takeoff mark(115) collinear to longer takeoff mark (187) will be used only when jets are landing simultaneous to launchings and therefore, while quickly launching all jets related to a mission, the ratio of aircraft taking of from aft and fore points will be 1:1.

The calculations for jets launching from fore point are more arduous since the aircraft will be depending heavily on its high AoA performance until it reaches V2min speed, and slight differential in assumption may end up giving huge variation in results.
Therefore, it makes sense to calculate a safe velocity from end of ski deck, that is derived directly from stall speed characterisitcs of basic flight of the same jet.

At 29 tons tow, with 26 ton thrust, considering 0 headwind and 10% allowance for Fr, Drag etc, the takeoff velocity will be 200 kmph and AoA of 15°. That would mean lift creating horizontal velocity of 190kmph and a throwup velocity of 50kmph.
The 50kmph is what has to be traded to buy time for creating extra lift.
 

Brumby

Major
Thats what they have calculated against, thats the max Mtow any jet is allowed from the aft launch point. Also i think the shorter takeoff mark(115) collinear to longer takeoff mark (187) will be used only when jets are landing simultaneous to launchings and therefore, while quickly launching all jets related to a mission, the ratio of aircraft taking of from aft and fore points will be 1:1.

The calculations for jets launching from fore point are more arduous since the aircraft will be depending heavily on its high AoA performance until it reaches V2min speed, and slight differential in assumption may end up giving huge variation in results.
Therefore, it makes sense to calculate a safe velocity from end of ski deck, that is derived directly from stall speed characterisitcs of basic flight of the same jet.

At 29 tons tow, with 26 ton thrust, considering 0 headwind and 10% allowance for Fr, Drag etc, the takeoff velocity will be 200 kmph and AoA of 15°. That would mean lift creating horizontal velocity of 190kmph and a throwup velocity of 50kmph.
The 50kmph is what has to be traded to buy time for creating extra lift.
Thanks.

At this stage I believe I have sufficient data to draw some conclusions.

The fore launch point i.e.; 115 m can only deliver a MTOW of between 24 to 25 tons. This essentially validates AFB's argument of 1/2 internal fuel load plus some basic weapons load out.

The aft launch point of 29 tons is delivered only through a T/W of 1.05. Is this a realistic position? Secondly the end speed of 200 kmh is way below the end speed reportedly of a cat launch at 145 KCAS i.e.270 kmh. As such I question whether 29 tons is actually achievable in practice.
 

by78

Brigadier
nice, but old pic - 3 numbers on the plane (102, as i spuspect)
What do you mean 'old' pic? Just because the plane is 'old', it doesn't mean every image of it is old. FYI, this is a screen capture from a Chinese TV segment from yesterday.
 

Bhurki

Junior Member
Registered Member
On which carrier have you ever seen such operation?
I understand navy does cyclic operational events and launching and recovery are non-concurrent.
But ,what else would be the use of having a collinear point half way down the longer launch position, other than putting it out of the right-of-way of landing aircraft?

The aft launch point of 29 tons is delivered only through a T/W of 1.05.
0.89* (26/29)
Thrust - 26t
Weight- 29t
Secondly the end speed of 200 kmh is way below the end speed reportedly of a cat launch at 145 KCAS i.e.270 kmh. As such I question whether 29 tons is actually achievable in practice.
According to the video launch, it certainly is.
You are forgetting about the vertical velocity that the ski deck provides to the jet that is not available in Cato.

Also the angle of attack directly after disconnect from Cat is a lot lower than ski deck provides.

This means that while a jet from Cat relies solely(almost) on aerodynamic lift ( dependent on 270kmph) for staying airborne, the ski jump jet also has the vertical component of engine thrust aiding it to be airborne( for 14°, it is 25% of total thrust) and also increased aero lift and force ( look ∆L/L, ∆F/F from your post)
 
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Intrepid

Senior Member
But ,what else would be the use of having a collinear point half way down the longer launch position, other than putting it out of the right-of-way of landing aircraft?
I don't know, may be for more space to handle aircraft.

Only position 3 is usable while aircraft are using the landing strip. Position 1 and 2 are not usable:

Liaoning.jpg
 

Bhurki

Junior Member
Registered Member
I don't know, may be for more space to handle aircraft.

Only position 3 is usable while aircraft are using the landing strip. Position 1 and 2 are not usable:

View attachment 56294
It probably provides about 20% extra sortie generation rate at the cost of ToW.
An ideal launch sequence of quickest time for putting a wings in the air will be 2-1-3, thereby letting pos 2 and 1 to reequip while 3 launches.
The kiev like launch positions would have provided much higher ToW with a lower SGR. Since the 001 is a lot wider, both the positions could've actually been equally pulled back to about 200m, albeit with loss of parking space in front of the island.
images (8).jpeg
 

Intrepid

Senior Member
An ideal launch sequence of quickest time for putting a wings in the air will be 2-1-3, thereby letting pos 2 and 1 to reequip while 3 launches.
I think only position 1 is normally used. And that's what the tire marks on the deck look like.
 

Brumby

Major
According to the video launch, it certainly is.
How do you actually calibrate the end speed by just watching a video? Can you please walk me through your calculations.

You are forgetting about the vertical velocity that the ski deck provides to the jet that is not available in Cato.
I have not forgotten the vertical velocity piece with a ski deck. The issue is to somehow equalize it to a cat launch i.e. apples to apples.

Also the angle of attack directly after disconnect from Cat is a lot lower than ski deck provides.

This means that while a jet from Cat relies solely(almost) on aerodynamic lift ( dependent on 270kmph) for staying airborne, the ski jump jet also has the vertical component of engine thrust aiding it to be airborne( for 14°, it is 25% of total thrust) and also increased aero lift and force ( look ∆L/L, ∆F/F from your post)
The end speed of a cat launch is reportedly around 145 KCAS. The MTOW is not specified nor whether a safety margin is already built into that speed. In the case of a cat launch, I am making the assumption that the end speed already provides sufficient velocity and lift to the aircraft once it leaves the deck. In the case of a ski deck launch I believe that is not the case.

The main contention is this. Your figures seem to suggest that a MTOW at 29 tons require a deck launch distance of 180 m but with an end speed of only 200 kmh. The argument being made is that the 200 kmh plus the vertical velocity is equivalent to a cat launch end speed of 270 kmh plus whatever safety margin speed that may be incorporated.

Is your position mathematically supportable or is it merely an assumption based on unknowns?
 
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